Author Topic: What is Swirl fraction in the hollow cone injection in Fluent?  (Read 2444 times)

Offline infocfd

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What is Swirl fraction in the hollow cone injection in Fluent?
« on: February 04, 2012, 06:32:06 PM »
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Hi,

How does the swirl fraction impact the hollow cone spray and how do I define it? Does anyone know?

Thank you.

Offline william

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Re: What is Swirl fraction in the hollow cone injection in Fluent?
« Reply #1 on: February 04, 2012, 06:33:24 PM »
In the hollow cone injection, the swirl fraction can be used to define the swirl velocity component, and then the spray cone angle. The swirl fraction is defined as

swirl_fraction = abs(tangential_velocity) / ( abs(tangential_velocity) + abs(axial_velocity) )

The coordinate system here is sitting on the rim of the injector with the velocity magnitude pointing in the direction of the axis that has been rotated by the cone angle and at the cone surface.

In this coordinate system, we will have u'=Vmag*(1-swirl_fraction), v'=0, w'=Vmag*swirl_fraction, where u' is the axial velocity component (injection direction), v' is the radial velocity component, w' is the tangential velocity component, and Vmag is the total velocity magnitude. If the swirl fraction is 0.5, then u'=0.5Vmag, v'=0, w'=0.5Vmag. However, the velocity vector has to be scaled to the unit vector to match the total velocity magnitude Vmag. Finally, we will have the velocity vector at the cone surface as

 u=(sqrt(2)/2)*Vmag, v=0, w=(sqrt(2)/2)*Vmag.

The swirl fraction is the relative fraction of the tangential velocity and the the other velocities at the cone surface. It is not the angle or the cosine of the angle. So, to get 45 degrees where the axial velocity and the tangential velocity would be equal, you would use a value of 0.5 for the swirl fraction. A value of zero would have no swirl, while a value of one would have no axial velocity component. It would look like a pinwheel. The swirling angle theta at the cone surface will be tan(theta) = swirl_fraction /(1-swirl_fraction).